This answer requires solving three subproblems:
How many bus arrivals $N$ do I expect to suffer through before experiencing a time between bus arrivals $> T_{min}$?
How long is the expected wait between buses $\mathbb{E}T_s$ given that $T_s \leq T_{min}$?
How long is the expected wait between two consecutive buses $\mathbb{E}T_l$ given that $T_l > T_{min}$?
The final answer is equal to $\mathbb{E}N \mathbb{E}T_s + \mathbb{E}T_l$, or perhaps just $\mathbb{E}N \mathbb{E}T_s$ if the time spent in the last interval isn't counted.
The answer to 3 is quite easy, thanks to the Exponential / Poisson assumption. The memoryless property of the Exponential distribution implies that $\mathbb{E}T_l = T_{min} + T_0$.
The answer to 2 is a little more involved; we have to find the expected value of the appropriate conditional random variate:
$$\frac{\int_0^{T_{\min}}xe^{-x/T_0}\text{d}x}{T_0(1-e^{-T_{min}/T_0})}$$
which, using integration by parts, solves to:
$$\frac{T_0 - (T_0+T_{min})e^{-T_{min}/T_0}}{1-e^{-T_{min}/T_0}}$$
As @Moormanly has observed, $N$ is distributed Geometric, in this case with parameter $p = \exp\{-T_{min}/T_0\}$. The expected value of a Geometric$(p)$ distribution is $(1-p)/p$, so, substituting and writing the full equation out, we get:
$$\left(\frac{1-e^{-T_{min}/T_0}}{e^{-T_{min}/T_0}}\right)\left(\frac{T_0 - (T_0+T_{min})e^{-T_{min}/T_0}}{1-e^{-T_{min}/T_0}}\right)+T_{min}+T_0$$
where the last two terms are optional depending upon the exact nature of the problem being solved. Making the obvious cancellation leads to:
$$\frac{T_0 - (T_0+T_{min})e^{-T_{min}/T_0}}{e^{-T_{min}/T_0}}+T_{min}+T_0$$
and some further algebra gets us to:
$$\frac{T_0}{e^{-T_{min}/T_0}}$$
Checking our results via simulation on your sample problem comes next. $T_0 = 10$ and $T_{min}=5$ gives us a calculated result of 16.483. Our simulation code:
res <- rep(0,10000)
for (i in 1:length(res)) {
res[i] <- x <- rexp(1,1/10)
while (x < 5) {
x <- rexp(1,1/10)
res[i] <- res[i] + x
}
}
c(mean(res), (mean(res)-10/exp(-0.5))/(sd(res)/sqrt(length(res))))
reports the simulated mean and the associated t-statistic for testing whether or not it is equal to the calculated value. The results are:
> c(mean(res), (mean(res)-10/exp(-0.5))/(sd(res)/sqrt(length(res))))
[1] 16.5793553 0.8998745
which would seem to confirm that we haven't messed up our algebra anywhere.